Show that the following equation has a solution in the interval (-1,1)












1












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Show that the following equation has a solution in the interval (-1,1)
I think its to do with intermediate value theorem but not sure what to do.
$$
frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1} = 0
$$






real-analysis







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    Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
    $endgroup$
    – Matti P.
    Dec 4 '18 at 11:42
















1












$begingroup$


Show that the following equation has a solution in the interval (-1,1)
I think its to do with intermediate value theorem but not sure what to do.
$$
frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1} = 0
$$






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
    $endgroup$
    – Matti P.
    Dec 4 '18 at 11:42














1












1








1


1



$begingroup$


Show that the following equation has a solution in the interval (-1,1)
I think its to do with intermediate value theorem but not sure what to do.
$$
frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1} = 0
$$






real-analysis







share|cite|improve this question











$endgroup$




Show that the following equation has a solution in the interval (-1,1)
I think its to do with intermediate value theorem but not sure what to do.
$$
frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1} = 0
$$






real-analysis




real-analysis






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edited Dec 4 '18 at 11:41









Arthur

113k7115197




113k7115197










asked Dec 4 '18 at 11:40









user607735user607735

103




103








  • 1




    $begingroup$
    Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
    $endgroup$
    – Matti P.
    Dec 4 '18 at 11:42














  • 1




    $begingroup$
    Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
    $endgroup$
    – Matti P.
    Dec 4 '18 at 11:42








1




1




$begingroup$
Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
$endgroup$
– Matti P.
Dec 4 '18 at 11:42




$begingroup$
Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
$endgroup$
– Matti P.
Dec 4 '18 at 11:42










2 Answers
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The limit as $x rightarrow -1^+$ (to the right of $-1$) gives you $+infty$ because of the term $frac{e^x - 0.25}{x + 1}$, you know that $e^{-1} - 0.25 > 0$ and hence the sign of $frac{e^x - 0.25}{x + 1}$ would depend on $x+1$ which is positive as $x rightarrow -1^+$. Similarly, the limit as $x rightarrow +1^-$ (to the left of $+1$) gives you $-infty$.



Imagine how the curve would look like, it has to cut the x-axis, unless your function is piecewise continuous, which is not your case.






share|cite|improve this answer











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    1












    $begingroup$

    If calculator is allowed, you can try several numbers from the interval.



    Consider the function $f(x)=frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1}, xin (-1,1)$. It is continuous in its domain.



    Now calculate:
    $$begin{align}f(-0.9)&=2.404>0 (f(0)=1.75>0 text{ will also do})\
    f(0.9)&=-3.433<0 end{align}$$
    By IVT, there must be a point $cin (-0.9,0.9)$, for which $f(c)=0$.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






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      active

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      active

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      3












      $begingroup$

      The limit as $x rightarrow -1^+$ (to the right of $-1$) gives you $+infty$ because of the term $frac{e^x - 0.25}{x + 1}$, you know that $e^{-1} - 0.25 > 0$ and hence the sign of $frac{e^x - 0.25}{x + 1}$ would depend on $x+1$ which is positive as $x rightarrow -1^+$. Similarly, the limit as $x rightarrow +1^-$ (to the left of $+1$) gives you $-infty$.



      Imagine how the curve would look like, it has to cut the x-axis, unless your function is piecewise continuous, which is not your case.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        The limit as $x rightarrow -1^+$ (to the right of $-1$) gives you $+infty$ because of the term $frac{e^x - 0.25}{x + 1}$, you know that $e^{-1} - 0.25 > 0$ and hence the sign of $frac{e^x - 0.25}{x + 1}$ would depend on $x+1$ which is positive as $x rightarrow -1^+$. Similarly, the limit as $x rightarrow +1^-$ (to the left of $+1$) gives you $-infty$.



        Imagine how the curve would look like, it has to cut the x-axis, unless your function is piecewise continuous, which is not your case.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          The limit as $x rightarrow -1^+$ (to the right of $-1$) gives you $+infty$ because of the term $frac{e^x - 0.25}{x + 1}$, you know that $e^{-1} - 0.25 > 0$ and hence the sign of $frac{e^x - 0.25}{x + 1}$ would depend on $x+1$ which is positive as $x rightarrow -1^+$. Similarly, the limit as $x rightarrow +1^-$ (to the left of $+1$) gives you $-infty$.



          Imagine how the curve would look like, it has to cut the x-axis, unless your function is piecewise continuous, which is not your case.






          share|cite|improve this answer











          $endgroup$



          The limit as $x rightarrow -1^+$ (to the right of $-1$) gives you $+infty$ because of the term $frac{e^x - 0.25}{x + 1}$, you know that $e^{-1} - 0.25 > 0$ and hence the sign of $frac{e^x - 0.25}{x + 1}$ would depend on $x+1$ which is positive as $x rightarrow -1^+$. Similarly, the limit as $x rightarrow +1^-$ (to the left of $+1$) gives you $-infty$.



          Imagine how the curve would look like, it has to cut the x-axis, unless your function is piecewise continuous, which is not your case.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 11:49

























          answered Dec 4 '18 at 11:44









          Ahmad BazziAhmad Bazzi

          8,0212724




          8,0212724























              1












              $begingroup$

              If calculator is allowed, you can try several numbers from the interval.



              Consider the function $f(x)=frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1}, xin (-1,1)$. It is continuous in its domain.



              Now calculate:
              $$begin{align}f(-0.9)&=2.404>0 (f(0)=1.75>0 text{ will also do})\
              f(0.9)&=-3.433<0 end{align}$$
              By IVT, there must be a point $cin (-0.9,0.9)$, for which $f(c)=0$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                If calculator is allowed, you can try several numbers from the interval.



                Consider the function $f(x)=frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1}, xin (-1,1)$. It is continuous in its domain.



                Now calculate:
                $$begin{align}f(-0.9)&=2.404>0 (f(0)=1.75>0 text{ will also do})\
                f(0.9)&=-3.433<0 end{align}$$
                By IVT, there must be a point $cin (-0.9,0.9)$, for which $f(c)=0$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If calculator is allowed, you can try several numbers from the interval.



                  Consider the function $f(x)=frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1}, xin (-1,1)$. It is continuous in its domain.



                  Now calculate:
                  $$begin{align}f(-0.9)&=2.404>0 (f(0)=1.75>0 text{ will also do})\
                  f(0.9)&=-3.433<0 end{align}$$
                  By IVT, there must be a point $cin (-0.9,0.9)$, for which $f(c)=0$.






                  share|cite|improve this answer









                  $endgroup$



                  If calculator is allowed, you can try several numbers from the interval.



                  Consider the function $f(x)=frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1}, xin (-1,1)$. It is continuous in its domain.



                  Now calculate:
                  $$begin{align}f(-0.9)&=2.404>0 (f(0)=1.75>0 text{ will also do})\
                  f(0.9)&=-3.433<0 end{align}$$
                  By IVT, there must be a point $cin (-0.9,0.9)$, for which $f(c)=0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 12:20









                  farruhotafarruhota

                  20k2738




                  20k2738






























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